A 0.240 M solution of the salt NaA has pH = 8.40. Calculate K_a for the acid HA. A) 2.63 × 10^-11 B) 3.80 × 10^-4 C) 1.05 × 10^-5 D) 6.60 × 10^-17 E) none of these

Question

Quizplus · Accepted Answer

Step 1: Write the chemical equation for the dissociation of NaA salt in water:

NaA → Na+ + A-

Step 2: Write the chemical equation for the dissociation of the acid HA in water:

HA + H2O ⇌ A- + H3O+

Step 3: Write the expression for the acidity constant (Ka) of the acid HA:

Ka = [A-][H3O+] / [HA]

Step 4: Since the solution is a buffer solution, the [A-] and [HA] concentrations are related by the Henderson-Hasselbalch equation:

pH = pKa + log([A-] / [HA])

8.40 = pKa + log([A-] / [HA])

Step 5: Solve for [A-] / [HA]:

[ A- ] / [HA] = 10^(pH - pKa)

[ A- ] / [HA] = 10^(-8.4 + pKa)

Step 6: Substitute the [A-] / [HA] ratio into the Ka expression:

Ka = [A-][H3O+] / [HA]

Ka = ([H3O+] * 10^(-8.4 + pKa)) / 1 M   (Assuming 1 M concentration of salt solution)

Step 7: Use the fact that in water, Kw = Ka * Kb (where Kw is the ionization constant of water) to solve for Kb of A- ion:

Kw = Ka * Kb

1.00 * 10^-14 = Ka * Kb

Kb = 1.00 * 10^-14 / Ka

Kb = 1.00 × 10S1U1P1-14 / (([H3O+] * 10^(-8.4 + pKa)) / 1 M)

Kb = 1.00 × 10S1U1P1-14 * (1 M) / ([H3O+] * 10^(-8.4 + pKa))

Kb = 1.00 × 10S1U1P1-14 * 10^(8.4 - pKa) / [H3O+]

Step 8: Convert Kb to pKb:

pKb = -log(Kb)

pKb = -log(1.00 × 10S1U1P1-14 * 10^(8.4 - pKa) / [H3O+])

pKb = -log(1.00 × 10S1U1P1-14) - (8.4 - pKa) + log([H3O+])

pKb = (pKa - 8.4) - log([H3O+]) + 14

pKb = (4.60) - 8.4 + log(2.40) + 14

pKb = 9.36

Step 9: Convert pKb to Kb:

Kb = 10^(-pKb)

Kb = 3.80 × 10S1U1P1-4

Therefore, the answer is B) 3.80 × 10S1U1P1-4.

A 0240 M Solution of the Salt NaA Has PH =